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\title[]{Jacobson's theorem for bilinear forms in characteristic 2} % This is the full title of the paper
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\author{Ahmed Laghribi}

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\begin{document}
\maketitle

\begin{abstract}
The aim of this note is to extend to bilinear forms in
characteristic $2$ a result of Jacobson which states that over any
field, two Albert quadratic forms are similar if and only if they
have the same Clifford invariant.
\end{abstract}

\begin{sloppypar}
\section{Introduction} Throughout this note $F$ denotes a field.
To a biquaternion algebras $Q_1\otimes_FQ_2$,  we attach a
$6$-dimesnional quadratic form $\phi$ given by $N_{Q_1}\perp
-N_{Q_2}\simeq {\bf H}\perp \phi$, where $N_{Q_i}$ is the norm
form of the quaternion algebra $Q_i$, and ${\bf H}$ is the
hyperbolic plane ($\perp$ and $\cong$ mean the orthogonal sum and
isometry, respectively). The form $\phi$ has trivial signed
discriminant (\resp trivial Arf invariant) if the characteristic
is $\neq 2$ (\resp if the characteristic is $2$). We call such a
form an Albert quadratic form. A well-known result of Jacobson
states that two biquaternion algebras are isomorphic if and only
if their corresponding Albert quadratic forms are similar
\cite{j}. In other words, this result says that two Albert
quadratic forms are similar if and only if they have the same
Clifford invariant. Using a method based on quadratic forms
theory, Mammone and Shapiro recovered Jacobson's result \cite{ms},
and also completed it in characteristic $2$ (see \cite[comments in
the middle of page 529]{ms}). The Clifford invariant of a
quadratic form (nonsingular quadratic form if the characteristic
is $2$) is defined in the $2$-torsion ${}_2\hskip-1mm\Br(F)$ of
the Brauer group of $F$. It is well-known that
${}_2\hskip-1mm\Br(F)$ is isomorphic to $I^2F/I^3F$ (\resp
$I^2_qF/I^3_qF$) if the characteristic is not $2$ (\resp if the
characteristic is $2$), where $I^kF=(IF)^k$,
$I^k_qF=I^{k-1}F\otimes W_q(F)$ and $IF$ denotes the ideal of the
Witt ring of even dimensional quadratic forms or bilinear forms
according as the characteristic is different or equal to $2$, and
$W_q(F)$ denotes the Witt group of nonsingualr quadratic forms
\cite{b1}. These isomorphisms are due to Merkurjev \cite{m} and
Sah \cite{s}, respectively. Since in characteristic $2$ we should
distinguish between quadratic and bilinear forms, it is natural to
ask whether an analogue of Jacobson's result holds for bilinear
forms in characteristic $2$. As for quadratic forms, an Albert
bilinear form means a $6$-dimensional form with trivial
determinant. Of course there is no notion of Clifford invariant
for bilinear forms in characteristic $2$, but we have a result of
Kato which gives an analogue of Merkurjev's and Sah's results
cited before (see Theorem \ref{t2}), and where the group of finite
sums of logarithmic differential forms (see below) is used as for
${}_2\hskip-1mm\Br(F)$ in the case of quadratic forms. We will see
that this ingredient suffices to get a result similar to that of
Jacobson.

Recall that for $n\geq 1$, one denotes by
$\Omega^n_F=\bigwedge^n\Omega^1_F$ the vector space of
$n$-differential forms over $F$, where $\Omega^1_F$ is the
$F$-vector space generated by symbols $dx$, $x\in F$, subject to
the relations: $d(x+y)=dx + dy$ and $d(xy)=xdy +ydx$, for $x,y\in
F$. An element $\frac{dx_1}{x_1}\wedge \cdots \wedge
\frac{dx_n}{x_n}$ is called an $n$-logarithmic symbol. A sum of
$n$-logarithmic symbols has length $k$ if it is a sum of $k$
$n$-logarithmic symbols but not a sum of less that $k$
$n$-logarithmic symbols.

For $a_1, \cdots, a_n\in F^*:=F\backslash \{0\}$, let $\left<a_1,
\cdots, a_n\right>_b$ denote the bilinear form given by
$\sum_{i=1}^na_ix_iy_i$. It is clear that any Albert bilinear form
$\gamma$ is isometric to $\alpha\left<r,s,rs, u,v,uv\right>_b$ for
suitable scalars $\alpha,r,s,u,v\in F^*$. We associate to such a
form $\gamma$ the following sum of $2$-logarithmic
symbols:\[e^2(\gamma):=\frac{dr}{r}\wedge \frac{ds}{s}+
\frac{du}{u}\wedge \frac{dv}{v}\in \Omega^2_F.\]

This is an invariant of $\gamma$ modulo $I^3F$ (see below). Our
main result in this note is the following theorem:

\begin{theorem} Let $F$ be a field of characteristic $2$, and let
$\gamma_1$, $\gamma_2$ be two Albert bilinear forms. Then we have
the following statements:\\(1) The length of $e^2(\gamma_1)$ is
$2$, $1$ or $0$ according as the Witt index of $\gamma_1$ is $0$,
$1$ or $3$.\\(2) $\gamma_1$ is similar to $\gamma_2$ if and only
if $e^2(\gamma_1)=e^2(\gamma_2)$. \label{t1}
\end{theorem}

\vskip2mm

By using Theorem \ref{t2} of Kato, we reduce the proof of Theorem
\ref{t1} to the use of methods from quadratic and bilinear forms
theory as was used in \cite{ms}. However, some of our results
differ from those given in \cite{ms}, and their proofs require
more details. The reason is that for bilinear forms in
characteristic $2$ some classical results, like the Witt
cancellation and the representation criterion, fail. Moreover, we
will be based on the connection between totally singular quadratic
forms and bilinear forms, and the notion of norm degree introduced
in \cite[Section 8]{hl1} will play a crucial r\^ole.

The rest of this note is organized as follows. We finish this
section by giving backgrounds on quadratic and bilinear forms in
characteristic $2$. The next section in devoted to some results
needed for the proof of Theorem \ref{t1}, and more specifically it
will concern the similarity of $4$-dimensional bilinear forms in
characteristic $2$, and then in the third section we prove the
theorem.

Form now on, we assume that $F$ is of characteristic $2$. The
expression ``bilinear form'' means ``regular symmetric bilinear
form of finite dimension''. To keep this note self-contained we
briefly recall some notions. More details can be found in
\cite{b1}, \cite{hl1}.

For a quadratic (or bilinear) form $\phi$, we denote by $\dim
\phi$ its dimension. Two quadratic (or bilinear) forms $\phi$ and
$\psi$ are called similar if $\phi\cong \alpha \psi$ for some
scalar $\alpha \in F^*$. A quadratic (or bilinear) form $\phi$ is
called a subform of another form $\psi$ if there exists a form
$\phi'$ such that $\psi\cong \phi \perp \phi'$.

To any bilinear form $B$ with underlying vector space $V$, we
associate a unique quadratic form $\widetilde{B}$ given by:
$\widetilde{B}(v)=B(v,v)$ for $v\in V$. A quadratic form $\phi$ is
called totally singular if $\phi\cong\widetilde{B}$ for some
bilinear form $B$. If $B\cong \left<a_1, \cdots, a_n\right>_b$,
then we denote $\widetilde{B}$ by $\left<a_1, \cdots, a_n\right>$.

For a field extension $K/F$ and a quadratic (or bilinear) form
$\phi$, the form $\phi\otimes K$ is denoted by $\phi_K$.

For a quadratic form $\phi$, we denote by $F(\phi)$ its function
field, i.e., the function field of the affine quadric given by
$\phi=0$. The function field of a bilinear form $B$ is by
definition the field $F(\widetilde{B})$.

A quadratic form $\phi$ with underlying vector space $V$ is
isotropic if there exists $v\in V\backslash\{0\}$ such that
$\phi(v)=0$. A bilinear form $B$ is isotropic if $\widetilde{B}$
is isotropic too. A form (quadratic or bilinear) is anisotropic if
it is not isotropic.

Any bilinear form $B$ decomposes as $B\cong B_{\an}\perp M_1\perp
\cdots \perp M_n$, where $B_{\an}$ is anisotropic and $M_i$ is
given by the matrix $\begin{pmatrix} a_i & 1 \\ 1& 0\end{pmatrix}$
for some $a_i\in F^*$ ($1\leq i\leq n$). The form $B_{\an}$ is
unique up to isometry, we call it the anisotropic part of $B$. The
interger $n$ is called the Witt index of $B$ and it is denoted by
$i_W(B)$. The form $B$ is called metabolic if $2i_W(B)=\dim B$.
Two bilinear forms $B$ and $C$ are called equivalent, denoted by
$B\sim C$, if $B\perp M\cong C\perp M'$ for some metabolic forms
$M$ and $M'$. By the uniqueness of the anisotropic part the
condition $B\sim C$ implies $B_{\an}\cong C_{\an}$. Recall that
the isotropy of a bilinear form $B$ is equivalent to say that
$i_W(B)\geq 1$.

The form $B=\left<1, a_1\right>_b\otimes \cdots \otimes
\left<1,a_n\right>_b$ is called an $n$-fold bilinear Pfister form,
and is denoted by $\left<\left<a_1,\cdots,a_n\right>\right>_b$. In
this case, we denote $\widetilde{B}$ by
$\left<\left<a_1,\cdots,a_n\right>\right>$ and we call it an
$n$-fold quasi-Pfister form. Recall that a bilinear Pfister form
is isotropic if and only if it is metabolic, and for any integer
$n\geq 1$, the ideal $I^nF$ is additively generated by $n$-fold
bilinear Pfister forms.

We say that a totally singular form $\phi$ is a quasi-Pfister
neighbor if there exists a quasi-Pfister form $\pi$ such that
$\phi$ is similar to a subform of $\pi$ and $2\dim \phi >\dim
\pi$. In this case, $\pi$ is unique up to isometry, and for any
field extension $K/F$, the forms $\phi_K$ and $\pi_K$ are
simultaneously isotropic or anisotropic.

The norm field of a nonzero totally singular form $\phi$, denoted
by $N_F(\phi)$, is the field $F^2(\alpha\beta \mid \alpha, \beta
\in D_F(\phi))$, where $D_F(\phi)$ is the set of scalars in $F^*$
represented by $\phi$. We denote by $\ndeg_F(\phi)$ the integer
$[N_F(\phi):F^2]$ and we call it the norm degree of $\phi$. It is
clear that $N_F(\phi)=N_F(\alpha\phi)$ for any scalar $\alpha\in
F^*$. If $\phi$ is anisotropic and $2^n<\dim \phi\leq 2^{n+1}$,
then $\ndeg_F(\phi)\geq 2^{n+1}$, and $\ndeg_F(\phi)=2^{n+1}$ if
and only if $\phi$ is a quasi-Pfister neighbor. If $\psi$ is a
quadratic form such that $\phi_{F(\psi)}$ is isotropic, then
$\psi$ is totally singular and $N_F(\psi)\subset N_F(\phi)$.
Moreover, there is a bijection between anisotropic $n$-fold
quasi-Pfister forms and purely inseparable extensions of $F^2$ of
degree $2^n$ inside $F$, it is given by $F^2(a_1, \cdots, a_n)
\leftrightarrow \left<\left<a_1, \cdots, a_n\right>\right>$. We
refer to \cite[Section 8]{hl1} for more details on norm field and
some of its applications.

\section{Preliminaries}
It is clear that the map $d:F\longrightarrow \Omega^1_F$:
$x\mapsto dx$, extends to a map $d:\Omega_F^n \lra \Omega_F^{n+1}$
defined by:\[d\left(x dx_1 \wedge dx_2 \wedge \cdots \wedge
dx_n\right)=dx \wedge dx_1 \wedge dx_2 \wedge \cdots \wedge
dx_n.\]There is a well-defined homomorphism $\wp_n: \Omega_F^n
\lra \Omega_F^n/d\Omega_F^{n-1}$ given on generators by:\[\wp_n
\left( x\frac{dx_{1}}{x_1}\wedge \cdots \wedge
\frac{dx_n}{x_n}\right)=\overline{(x^2-x)\frac {dx_1}{x_1}\wedge
\cdots \wedge \frac{dx_n}{x_n}}.\]

We write $\nu_F(n)$ the kernel of this map. A crucial result that
we will use is the following theorem due to Kato which gives a
link between this kernel and bilinear forms. Kato also established
a relation between the cokernel of $\wp_n$ and quadratic forms,
but we don't need it here.

\begin{theorem} [\cite{k}] For any integer $n\geq 1$, there is an isomorphism
$e^n:I^nF/I^{n+1}F\stackrel{\sim}\lra \nu_F(n)$, given
by:\[e^n\left(\sum \left<\left<a_1, \cdots,
a_n\right>\right>_b+I^{n+1}F\right)=\sum \frac{da_1}{a_1}\wedge
\cdots \wedge\frac{da_n}{a_n}.\] \label{t2}
\end{theorem}

We will use this theorem in the case $n=2$. Another result that we
need is the following theorem which gives information on the
dimensions of bilinear forms in $I^nF$:\vskip2mm

\begin{theorem}  Let $B\in I^nF$ be anisotropic ($n\geq 1$).
Then:\\(1) (\cite{l2}) $\dim B=0$ or $\dim B\geq 2^n$.\\(2) If
$\dim B>2^n$, then $\dim B\geq 2^n+2^{n-1}$. \label{t3}
\end{theorem}
\vskip2mm Statement (2) can be deduced from a result of Vishik
\cite{v} by the same argument used for the proof of \cite[Prop.
5.7]{l2}.\vskip2mm

\begin{lemma} Let $B$ be an anisotropic bilinear form and $d\in
F^*\backslash F^{*2}$. Then, $B$ becomes metabolic over
$F(\sqrt{d})$ if and only if $B\cong \alpha_1\left<1,
d+x_1^2\right>_b\perp \cdots \perp \alpha_n\left<1,
d+x_n^2\right>_b$ for some $\alpha_i, x_i\in F$ with $\alpha_i\neq
0$, $1\leq i\leq n$. \label{lem1}
\end{lemma}

\begin{proof} Use \cite[Lem. 3.4]{l1} and the fact that for $a\neq 0,
b\in F$, the bilinear form given by the matrix $\begin{pmatrix} a&
b\\ b& ad\end{pmatrix}$ is isometric to $\alpha\left< 1,
d+x^2\right>_b$ for some $\alpha\neq 0, x\in F$.\end{proof}

Some results on bilinear forms of dimension $4$ will be needed,
more particularly properties on the similarity between such forms.
Before we give our contribution in this direction (Proposition
\ref{p1} and Corollary \ref{cor1}), we start by clarifying the
situation whether a $4$-dimensional bilinear form becomes
isotropic over the inseparable quadratic extension given by its
determinant. Recall that an anisotropic quadratic form of
dimension $4$ (nonsingular if the characteristic is $2$) stays
anisotropic over the quadratic extension given by its signed
discriminant (or the separable quadratic extension given by its
Arf invariant). For bilinear forms in characteristic $2$ the
situation is different as shows the following
proposition:\vskip2mm

\begin{proposition}
Let $B$ be a bilinear form of dimension $4$ whose determinant is
not trivial. Then, $B$ becomes isotropic over the quadratic
inseparable extension given by its determinant if and only if
$\ndeg_F(\widetilde{B})\leq 4$. \label{p3}
\end{proposition}

\begin{proof} Write $B=\alpha \left< r,s,rs, d\right>_b$ for suitable scalars
$\alpha, r,s\in F^*$. One has $N_F(\widetilde{B})=F^2(r,s,d)$ and
thus $\ndeg_F(\widetilde{B})\leq 8$. We may suppose that $B$ is
anisotropic, in particular $\left<r,s,rs\right>_b$ is anisotropic
too, and thus $[F^2(r,s):F^2]=4$. Hence,
$\ndeg_F(\widetilde{B})\in \{4, 8\}$.

Suppose $\ndeg_F(\widetilde{B})=4$. Hence, $d\in F^2(r,s)$. Since
$d\not\in F^2$, one can write $F^2(r,s)=F^2(d,k)$ for some $k\in
F^*$. In particular, $\left<\left<r,s\right>\right>\cong
\left<\left<d,k\right>\right>$. Now it is clear that
$\left<\left<r,s\right>\right>_{F(\sqrt{d})}$ is isotropic. Then,
$\left<r,s,rs, d\right>_{F(\sqrt{d})}$ is also isotropic, i.e.,
$B_{F(\sqrt{d})}$ is isotropic.

Conversely, if $B_{F(\sqrt{d})}$ is isotropic, then
$(\left<\left<r,s\right>\right>_b)_{F(\sqrt{d})}$ is metabolic. By
Lemma \ref{lem1} it is clear $\left<\left<r,s\right>\right>_b
\cong \left<\left<d+x^2,y\right>\right>$ for suitable scalars
$x,y\in F$. Hence, $d\in F^2(r,s)$ and then
$\ndeg_F(\widetilde{B})=4$.\end{proof}

The following proposition is in the spirit of a result due to
Wadsworth \cite[Theorem 7]{w}. In our case, the notion of norm
degree plays an essential r\^ole:\vskip2mm

\begin{proposition}
Let $B=\left< r,s,rs,d\right>_b$ and $C=\left< u,v,uv,d\right>_b$
be two anisotropic bilinear forms of dimension $4$ having the same
determinant $d$. Suppose that $\ndeg_F (\widetilde{B})=8$ and
$\left< r,s,rs,u,v,uv\right>$ is isotropic. Then, $B$ and $C$ are
similar over $F(\sqrt{d})$ if and only if there exists $x\in F^*$
such that $\left< r,s,rs,d+x^2\right>$ is similar to
$\left<u,v,uv,d+x^2\right>$. \label{p1}
\end{proposition}

\begin{proof} It is clear that $N_F(\widetilde{B})=F^2(r,s,d)$. Suppose that $\left< r,s,rs,d+x^2\right>_b$ is
similar to $\left<u,v,uv,d+x^2\right>_b$ for some $x\in F$. Since
$(\left<d+x^2\right>_b)_{F(\sqrt{d})}\cong
(\left<1\right>_b)_{F(\sqrt{d})}\cong
(\left<d\right>_b)_{F(\sqrt{d})}$, it is clear that the forms
$\left< r,s,rs,d\right>_b$ and $\left<u,v,uv,d\right>_b$ are
similar over $F(\sqrt{d})$.

Conversely, suppose that $B_{F(\sqrt{d})}$ is similar to
$C_{F(\sqrt{d})}$. Then, by using the multiplicativity of bilinear
Pfister forms, we get that $(\left<\left<r,s\right>\right>_b\perp
\left<\left<u,v\right>\right>_b)_{F(\sqrt{d})}$ is metabolic. The
isotropy of $\left< r,s,rs,u,v,uv\right>_b$ and Lemma \ref{lem1}
imply that\[\left<\left<r,s\right>\right>_b\perp
\left<\left<u,v\right>\right>_b\sim \alpha\left<1, d+x^2\right>_b
\perp \beta\left<1, d+y^2\right>_b\]for suitable $\alpha, \beta,
x,y\in F$ with $\alpha, \beta\neq 0$. But by comparing
determinants in the last relation, we may suppose that $x=y$.
Moreover, by the subform theorem for bilinear forms \cite[Prop.
1.1]{l1}, we may suppose that $\alpha$ is represented by $\left<
r,s, rs\right>_b$, and thus $\left<\left< r,s\right>\right>_b\cong
\alpha\left<\left< r,s\right>\right>_b$. Hence we
get\[\alpha(\left<\left<r,s\right>\right>_b\perp \left<1,
d+x^2\right>_b)\sim  \left<\left<u,v\right>\right>_b\perp
\beta\left<1, d+x^2\right>_b.\]The anisotropic part of
$\left<\left<r,s\right>\right>_b\perp \left<1, d+x^2\right>_b$ has
dimension $4$, and thus it is isometric to $\left<r,s, rs,
d+x^2\right>_b$, otherwise the form $\left< r,s,rs\right>_b$ would
represent $d+x^2$, and thus $d\in F^2(r,s)$, a contradiction with
$\ndeg_F(\widetilde{B})=8$. By the uniqueness of the anisotropic
part, the form $\left<\left<u,v\right>\right>_b\perp \beta\left<1,
d+x^2\right>_b$ also has an anisotropic part of dimension $4$. It
follows from the multiplicativity of bilinear Pfister forms that
$\alpha\left<r,s, rs, d+x^2\right>_b\cong \lambda\left<u,v, uv,
d+x^2\right>_b$ for some scalar $\lambda \in F^*$. Hence, the
claim.\end{proof}

We don't know if Proposition \ref{p1} remains true in the case of
norm degree $4$. As a corollary of this proposition we get the
following:\vskip2mm

\begin{corollary} Let $B$ and
$C$ be anisotropic bilinear forms of dimension $4$ such that
$B\perp C\in I^3F$. Then, $B$ is similar to $C$. \label{cor1}
\end{corollary}

\begin{proof} The forms $B$ and $C$ have
the same determinant since $B\perp C\in I^3F$. Set
$B=\alpha\left<r,s,rs,l\right>_b$ and
$C=\beta\left<u,v,uv,l\right>_b$. We may suppose that $l$ is not a
square, otherwise we get the similarity by Theorem \ref{t3} and
the multiplicativity of bilinear Pfister forms.

Since $B\perp C\in I^3F$, it follows from Theorem \ref{t2} that
$e^2(B\perp C)=\frac{du}{u}\wedge\frac{dv}{v}+
\frac{dr}{r}\wedge\frac{ds}{s}+
\frac{d(\alpha\beta)}{\alpha\beta}\wedge\frac{dl}{l}=0$. Hence, by
statement (1) of Theorem \ref{t1} the form
$\left<r,s,rs,u,v,uv\right>_b$ is isotropic (note that we can use
statement (1) of Theorem \ref{t1} since its proof is independent
of this corollary). Moreover, Theorem \ref{t3}(1) implies that the
forms $B_{F(C)}$ and $C_{F(B)}$ are isotropic, and thus
$N_F(\widetilde{B})=N_F(\widetilde{C})=F^2(r,s,l)$. Since $B\perp
C\in I^3F$, we get $(\left<\left<r,s\right>\right>_b\perp
\left<\left<u,v\right>\right>_b)_{F(\sqrt{l})}\in I^3F(\sqrt{l})$,
and again by Theorem \ref{t3}(1) we deduce the isometry
$(\left<\left<r,s\right>\right>_b)_{F(\sqrt{l})}\cong
(\left<\left<u,v\right>\right>_b)_{F(\sqrt{l})}$. Now it is clear
that $B_{F(\sqrt{l})}$ similar to $C_{F(\sqrt{l})}$. We discuss
two cases:

(1) Suppose that $\ndeg_F(\widetilde{B})=8$: In this case we
conclude by Proposition \ref{p1} that
$\left<r,s,rs,l+x^2\right>_b\cong k\left<u,v,uv,l+x^2\right>_b$
for some scalars $x,k\neq 0\in F$. In
particular,\[\left<\left<r,s\right>\right>_b\sim \left<1,
l+x^2\right>_b\perp k\left<\left<u,v\right>\right>_b \perp
k\left<1, l+x^2\right>_b.\]If we combine this relation with
$B\perp C\in I^3F$, it is clear that modulo $I^3F$ we get\[\alpha
k\left< 1, l+x^2\right>_b \perp \alpha\left< l, l+x^2\right>_b
\perp \beta\left< 1, l\right>_b \in I^3F.\]Hence\[\alpha k\left<
1, l+x^2\right>_b \perp \alpha\left< l, l+x^2\right>_b \sim
\beta\left< 1, l\right>_b.\]Since $\left<r,s,rs,l\right>_b \sim
k\left<\left<u,v\right>\right>_b \perp k\left<1,  l+x^2\right>_b
\perp \left< l, l+x^2\right>_b$, it follows that
$\left<r,s,rs,l\right>_b \sim k\left<\left<u,v\right>\right>_b
\perp \alpha\beta\left< 1,l\right>_b$. By the uniqueness of the
anisotropic part and the multiplicativity of bilinear Pfister
forms, we conclude that $B\cong mC$ for some scalar $m\in F^*$.

(2) Suppose that $\ndeg_F(\widetilde{B})=4$: Since $\left<r,s,
rs\right>_b$ is anisotropic, the form
$\left<\left<r,s\right>\right>_b$ is anisotropic too, and then
$[F^2(r,s):F^2]=4$. Hence,
$N_F(\widetilde{B})=N_F(\widetilde{C})=F^2(r,s)=F^2(u, v)$ and
$l\in F^2(r,s)$.

(i) If $B\perp C$ is isotropic, then it is metabolic and thus
$B\cong C$.

(ii) If $B\perp C$ is anisotropic. By Theorem \ref{t3}(1), $B\perp
C$ becomes metabolic over its function field. It follows from
\cite[Cor. 5.5]{l1} that $B\perp C$ is similar to $3$-fold
bilinear Pfister form. Hence, $\ndeg_F(\widetilde{B\perp
C})=8=\ndeg_F(\alpha(\widetilde{B\perp C}))$. By a simple
computation with the fact $l\in F^2(r,s)=F^2(u, v)$, one has
$N_F(\alpha(\widetilde{B\perp C}))=F^2(r,s,\alpha\beta)$.
Moreover, $\left<\left<r,s\right>\right>\cong
\left<\left<u,v\right>\right>$ since $F^2(r,s)=F^2(u, v)$. Hence,
$\left<u,v,uv\right>$ is isotropic over
$K:=F(\left<\left<r,s\right>\right>)$ since it is a quasi-Pfister
neighbor of $\left<\left<u,v\right>\right>$. Consequently,
$(\left<u,v,uv\right>_b)_K \sim (\left< 1\right>_b)_K$ and
similarly $(\left<r,s,rs\right>_b)_K \sim (\left< 1\right>_b)_K$.
Hence, after extending the relation $\alpha(B\perp C)\in I^3F$ to
the field $K$, we conclude that $(\left<\left< l,
\alpha\beta\right>\right>_b)_{K}$ is metabolic. In particular,
$\left<\left< l, \alpha\beta\right>\right>\cong
\left<\left<e,f\right>\right>$ for some $e,f\in F^2(r,s)$
\cite[Th. 1.2]{l1}. This implies that $\alpha\beta \in F^2(r,s)$,
and thus $[F^2(r,s,\alpha\beta):F^2]=4$, a
contradiction.\end{proof}

\section{Proof of Theorem \ref{t1}}

(1) (i) If $e^2(\gamma_1)$ has length $2$, then it is clear that
$i_W(\gamma_1)=0$.

(ii) If $e^2(\gamma_1)$ has length $1$: Let $k,l\in F^*$ be such
that $e^2(\gamma_1)=\frac{dk}{k}\wedge \frac{dl}{l}$. Let $K$ be
the function field of $\tau=\left<\left<k,l\right>\right>_b$.
Since $\tau_K$ is metabolic, one has $e^2(\gamma_1)_K=0$. It
follows from Theorem \ref{t2} that $(\gamma_1)_K\in I^3K$, and by
Theorem \ref{t3} $(\gamma_1)_K$ is metabolic. Since
$\ndeg_F(\widetilde{\tau})=4$ because $\tau$ is anisotropic, it
follows from \cite[Th. 1.2]{l1} that $\gamma_1$ is isotropic.
Moreover, the form $\gamma_1$ is not metabolic by reason of
lenght, hence $i_W(\gamma_1)=1$.

(iii) If $e^2(\gamma_1)$ has length $0$, then it follows from
Theorems \ref{t2} and \ref{t3} that $i_W(\gamma_1)=3$.

\vskip2mm

(2) Let $\gamma_1$, $\gamma_2$ be two Albert bilinear forms. We
have to show that $\gamma_1$ is similar to $\gamma_2$ if and only
if $e^2(\gamma_1)=e^2(\gamma_2)$.

Suppose that $\gamma_1$ is similar to $\gamma_2$. Then, $\gamma_1
\perp \gamma_2\in I^3F$. Il follows from Theorem \ref{t2} that
$e^2(\gamma_1)=e^2(\gamma_2)$.

Conversely, suppose that $e^2(\gamma_1)=e^2(\gamma_2)$. Then,
again by Theorem \ref{t2} $\gamma_1 \perp \gamma_2\in I^3F$. After
multiplying, if necessary, $\gamma_1$ and $\gamma_2$ by suitable
scalars, we may suppose that $\gamma_1 \perp \gamma_2$ is
isotropic. By Theorem \ref{t3}(2) the Witt index of $\gamma_1\perp
\gamma_2$ is at least $2$. Let $k,l\in F^*$ be such that
$\gamma_1\cong k\left< 1, l\right> \perp B$ and $\gamma_2=k\left<
1, l\right>\perp C$ for some $4$-dimensional bilinear forms $B$
and $C$ which have the same determinant $l$. Write $B=\alpha\left<
r,s,rs, l\right>$ and $C=\beta\left<u,v,uv, l\right>$. An easy
computation of $e^2(\gamma_1)= e^2(\gamma_2)$ gives
that\[\frac{dr}{r}\wedge\frac{ds}{s} +
\frac{dl}{l}\wedge\frac{d(k\alpha)}{k\alpha}=
\frac{du}{u}\wedge\frac{dv}{v} +
\frac{dl}{l}\wedge\frac{d(k\beta)}{k\beta}.\]In
particular,\[\frac{dr}{r}\wedge\frac{ds}{s} +
\frac{du}{u}\wedge\frac{dv}{v}=
\frac{dl}{l}\wedge\frac{d(k^2\alpha\beta)}{k^2\alpha\beta}.\]We
conclude by statement (1) that $\left<r,s,rs,u,v,uv\right>_b$ is
isotropic. Since $B\perp C\in I^3F$, it follows from Corollary
\ref{cor1} that $B\cong mC$ for some scalar $m\in F^*$. Since
$\gamma_1\perp \gamma_2\in I^3F$, one deduces that $m\gamma_1\perp
\gamma_2\in I^3F$, i.e., $mk\left<1, l\right>\perp k\left< 1,
l\right>\in I^3F$. Hence, $mk\left<1, l\right>\cong k\left< 1,
l\right>$, and $\gamma_1\cong m\gamma_2$.


\begin{thebibliography}{99}

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{\bibname A. Laghribi}, `Sur le d\'eploiement des formes
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%To appear in {\em Pacific J. Math.}

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\end{thebibliography}

\affiliationone{% in this example, two authors share an institution
   Ahmed Laghribi\\
   Facult\'e Jean Perrin\\Laboratoire de Math\'ematiques de Lens\\Rue Jean Souvraz - SP18\\F-62307 Lens
   \email{laghribi@euler.univ-artois.fr}}

\end{sloppypar}

\end{document}

